1. The problem gives a frequency distribution table with intervals of hours per day (Horas/dia), frequencies $n_i$, and cumulative relative frequencies $Fr_i$. 2. We are given partial values $a$ and $b$ and asked to find which pair $(a,b)$ fits the data. 3. The cumulative relative frequency $Fr_i$ is calculated as $Fr_i = \frac{\sum n_i}{N}$ where $N$ is the total number of observations. 4. From the table: - For interval [1; 2[, $n_1=2$, $Fr_1=0.10$ - For interval [2; 3[, $n_2=5$, $Fr_2=b$ - For interval [3; 4[, $n_3=a$, $Fr_3=0.85$ - For interval [4; 5[, $n_4=3$, $Fr_4=1$ 5. Total observations $N = 2 + 5 + a + 3 = 10 + a$ 6. Calculate cumulative frequencies: - $Fr_1 = \frac{2}{N} = 0.10 \Rightarrow 2 = 0.10N \Rightarrow N = 20$ 7. Using $N=20$, find $a$: - $N = 10 + a = 20 \Rightarrow a = 10$ 8. Calculate $Fr_2 = \frac{2 + 5}{20} = \frac{7}{20} = 0.35$, so $b=0.35$ 9. Check $Fr_3 = \frac{2 + 5 + 10}{20} = \frac{17}{20} = 0.85$ which matches the table. 10. Check $Fr_4 = 1$ since total frequency sums to $20$. 11. Therefore, the correct pair is $a=10$ and $b=0.35$. **Final answer:** $a=10$ and $b=0.35$ corresponds to option (C).